Member Avatar for tazboy

list is supposed to be an array of pointers, so hopefully I declared that correctly.
This function is supposed to read lines of text and each line of text should have a pointer pointing to it so that I can sort them.

What I don't understand is when I read in the text do I have to save each line to an array and then my array of pointers would point to that array? I'm thinking each line should be in an array because then I could print out the whole line afterward. Is that the best way to do it?

char *read_file( istream *infile, data_t *lines_ptr ){
 
    char *list = new char[LIST_SIZE];
    char ch;
    int line = 0;
    
    // starts out true so that the first line of text is stored
    bool char_after_newline = true;
    
    while ( (ch = infile->get()) != EOF ){
	
	cout << ch;
	
	// Executes if there is a new line and then fills an array with
	// the characters 
	if ( char_after_newline == true ){
	    
	    // NEED HELP HERE
	    line++;
	}
	
	// This if-else statement marks whether a new line is coming up
	if ( ch == '\n' ){
	    
	    char_after_newline = true;
	    
	} else {
	    
	    char_after_newline = false;
	    
	}

    }
    
    // Save the number of lines for use in the calling function
    *lines_ptr = line;

    return list;
    
}

Thanks for any help.

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Member Avatar for Ancient Dragon

>>list is supposed to be an array of pointers, so hopefully I declared that correctly.

No, you did not declare it correctly. What you declared was an array of characters. Here is an array of pointers: char *list[LIST_SIZE]; If you want each line of the file to be in the array

string line;
int count = 0;
while( getline(infile, line) )
{
    list[count] = strdup(line.c_str());
    ++count;
}

Since this is c++, not C, it would be even easier to use an array of std::string's or a std::vector

string list[LIST_MAX];
string line;
int count = 0;
while( getline(infile, line) )
{
    list[count] = line;
    ++count;
}
Member Avatar for tazboy

char *list[LIST_SIZE];

Without vectors, I need it to be a dynamic array using new. So would it be: char **list = new char[LIST_SIZE]; or char **list = new *char[LIST_SIZE]; If this is true then I would need to have my function return char **read_line , correct?

Member Avatar for VernonDozier

Without vectors, I need it to be a dynamic array using new. So would it be: char **list = new char[LIST_SIZE]; or char **list = new *char[LIST_SIZE]; If this is true then I would need to have my function return char **read_line , correct?

Is list type char* or type char** ? You have this in your code:

char *list = new char[LIST_SIZE];

That would make list type char*. If list is to be an array of pointers, it should be type char**. If list is type char**, both of these would be incorrect:

char **list = new char[LIST_SIZE];
char **list = new *char[LIST_SIZE];

This would be correct:

char **list = new char*[LIST_SIZE];

Then if you wanted to have each char* be a character array, you could now set aside space using new again:

char **list = new char*[LIST_SIZE];
for (int i = 0; i < LIST_SIZE; i++)
{
     list[i] = new char[50];
}

This is assuming you want an array of LIST_SIZE C-Style strings, each array element having a capacity of 50 characters.

Member Avatar for VernonDozier

Here's a sample program that may be useful:

#include <iostream>
using namespace std;

int main()
{
    int LIST_SIZE, NUM_CHARS_PER_ELEMENT;
    cout << "Enter number of elements in list: ";
    cin >> LIST_SIZE;
    cout << "Enter number of characters in each array element: ";
    cin >> NUM_CHARS_PER_ELEMENT;
    
    char **list = new char*[LIST_SIZE];
    for (int i = 0; i < LIST_SIZE; i++) 
    {
        list[i] = new char[NUM_CHARS_PER_ELEMENT];
    }
    
    cout << "Enter " << LIST_SIZE << " names\n";
    for (int i = 0; i < LIST_SIZE; i++)
    {
        cout << "Enter name " << i << " : ";
        cin >> list[i];
    }
    
    cout << "Here are the names\n";
    for (int i = 0; i < LIST_SIZE; i++)
    {
        cout << "Name " << i << " : " << list[i] << endl;
    }

    return 0;
}
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